Chapter 1 Matter, Measurement and Problem Solving

Chemistry: A Molecular Approach, 1st Ed.
Nivaldo Tro
Chapter 1
Matter,
Measurement,
and Problem
Solving
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2008, Prentice Hall
Composition of Matter
Atoms and Molecules
Scientific Method
Structure Determines Properties
• the properties of matter are determined by the atoms
and molecules that compose it
1.
2.
3.
4.
carbon monoxide
composed of one carbon atom
and one oxygen atom
colorless, odorless gas
burns with a blue flame
binds to hemoglobin
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1.
2.
3.
4.
carbon dioxide
composed of one carbon atom
and two oxygen atoms
colorless, odorless gas
incombustible
does not bind to hemoglobin
3
Atoms and Molecules
• atoms
 are submicroscopic particles
 are the fundamental building blocks of all matter
• molecules
 two or more atoms attached together
 attachments are called bonds
 attachments come in different strengths
 molecules come in different shapes and patterns
• Chemistry is the science that seeks to understand the
behavior of matter by studying the behavior of atoms
and molecules
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The Scientific Approach to Knowledge
• philosophers try to understand the universe by
reasoning and thinking about “ideal” behavior
• scientists try to understand the universe through
empirical knowledge gained through observation
and experiment
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From Observation to Understanding
• Hypothesis – a tentative interpretation or explanation
for an observation
 falsifiable – confirmed or refuted by other observations
 tested by experiments – validated or invalidated
• when similar observations are consistently made, it can
lead to a Scientific Law
 a statement of a behavior that is always observed
 summarizes past observations and predicts future ones
 Law of Conservation of Mass
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From Specific to General Understanding
• a hypothesis is a potential explanation for a
single or small number of observations
• a theory is a general explanation for the
manifestation and behavior of all nature
models
pinnacle of scientific knowledge
validated or invalidated by experiment and
observation
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a test of a
hypothesis
or theory
Scientific Method
the careful noting
and recording of
natural phenomena
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a tentative explanation of a
single or small number of
natural phenomena
a general explanation
of natural phenomena
a generally observed
natural phenomenon
8
Classification of Matter
States of Matter
Physical and Chemical Properties
Physical and Chemical Changes
Classification of Matter
• matter is anything that has mass and occupies
space
• we can classify matter based on whether it’s
solid, liquid, or gas
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Classifying Matter
by Physical State
• matter can be classified as solid, liquid, or gas
based on the characteristics it exhibits
State
Shape
Volume
Compress
Flow
Solid
Fixed
Fixed
No
No
Liquid
Indef.
Fixed
No
Yes
Gas
Indef.
Indef.
Yes
Yes
• Fixed = keeps shape when placed in a container
• Indefinite = takes the shape of the container
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Solids
• the particles in a solid are packed close
together and are fixed in position
 though they may vibrate
• the close packing of the particles results
•
in solids being incompressible
the inability of the particles to move
around results in solids retaining their
shape and volume when placed in a new
container, and prevents the particles
from flowing
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Crystalline Solids
• some solids have their
particles arranged in an
orderly geometric pattern –
we call these crystalline
solids
salt and diamonds
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Amorphous Solids
• some solids have their
particles randomly
distributed without any
long-range pattern – we call
these amorphous solids
plastic
glass
charcoal
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Liquids
• the particles in a liquid are closely
•
•
packed, but they have some ability to
move around
the close packing results in liquids
being incompressible
but the ability of the particles to
move allows liquids to take the shape
of their container and to flow –
however, they don’t have enough
freedom to escape and expand to fill
the container
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Gases
• in the gas state, the particles
•
•
have complete freedom from
each other
the particles are constantly
flying around, bumping into
each other and the container
in the gas state, there is a lot of
empty space between the
particles
 on average
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Gases
• because there is a lot of
empty space, the particles
can be squeezed closer
together – therefore gases
are compressible
• because the particles are not
held in close contact and are
moving freely, gases expand
to fill and take the shape of
their container, and will flow
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Classification of Matter
by Composition
• matter whose composition does not change from one
sample to another is called a pure substance
 made of a single type of atom or molecule
 because composition is always the same, all samples have
the same characteristics
• matter whose composition may vary from one sample
to another is called a mixture
 two or more types of atoms or molecules combined in
variable proportions
 because composition varies, samples have the different
characteristics
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Classification of Matter
by Composition
1) made of one type of
particle
2) all samples show the
same intensive
properties
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1) made of multiple
types of particles
2) samples may show
different intensive
properties
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Classification of Pure Substances
• substances that cannot be broken down into simpler
substances by chemical reactions are called elements
 basic building blocks of matter
 composed of single type of atom
 though those atoms may or may not be combined into molecules
• substances that can be decomposed are called compounds
 chemical combinations of elements
 composed of molecules that contain two or more different kinds
of atoms
 all molecules of a compound are identical, so all samples of a
compound behave the same way
• most natural pure substances are compounds
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Classification of Pure Substances
1) made of one
type of atom
(some elements
found as multiatom
molecules in
nature)
2) combine
together to
make
compounds
1) made of one
type of
molecule, or
array of ions
2) molecules
contain 2 or
more different
kinds of atoms
21
Classification of Mixtures
• homogeneous = mixture that has uniform
composition throughout
 every piece of a sample has identical characteristics, though
another sample with the same components may have
different characteristics
 atoms or molecules mixed uniformly
• heterogeneous = mixture that does not have uniform
composition throughout
 contains regions within the sample with different
characteristics
 atoms or molecules not mixed uniformly
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Classification of Mixtures
1) made of
multiple
substances,
whose
presence can
be seen
2) portions of a
sample have
different
composition
and properties
1) made of
multiple
substances, but
appears to be
one substance
2) all portions of
a sample have
the same
composition
and properties
23
Separation of Mixtures
• separate mixtures based on different
physical properties of the components
Physical change
Different Physical Property
Technique
Boiling Point
Distillation
State of Matter (solid/liquid/gas)
Filtration
Adherence to a Surface
Chromatography
Volatility
Evaporation
Density
Centrifugation &
Decanting
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Distillation
25
Filtration
26
Changes in Matter
• changes that alter the state or appearance of the
matter without altering the composition are
called physical changes
• changes that alter the composition of the matter
are called chemical changes
during the chemical change, the atoms that are
present rearrange into new molecules, but all of the
original atoms are still present
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Physical Changes in Matter
The boiling of
water is a physical
change. The water
molecules are
separated from
each other, but
their structure and
composition do
not change.
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Chemical Changes in Matter
The rusting of iron
is a chemical
change. The iron
atoms in the nail
combine with
oxygen atoms
from O2 in the air
to make a new
substance, rust,
with a different
composition.
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Properties of Matter
• physical properties are the characteristics of
matter that can be changed without changing
its composition
characteristics that are directly observable
• chemical properties are the characteristics
that determine how the composition of matter
changes as a result of contact with other matter
or the influence of energy
characteristics that describe the behavior of matter
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Common Physical Changes
• processes that cause
changes in the matter
that do not change its
composition
• state changes
Dissolving
Subliming of Sugar
Dry Ice
C12H22O11(s)
Dry Ice
boiling / condensing
melting / freezing
subliming
• dissolving
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CO2(g)
CO2(s)
C12H22O11(aq)
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Common Chemical Changes
• processes that cause
changes in the matter
that change its
composition
• rusting
• processes that release
lots of energy
• burning
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
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Energy
Energy Changes in Matter
• changes in matter, both physical and chemical, result
•
•
in the matter either gaining or releasing energy
energy is the capacity to do work
work is the action of a force applied across a distance
 a force is a push or a pull on an object
 electrostatic force is the push or pull on objects that have an
electrical charge
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Energy of Matter
• all matter possesses energy
• energy is classified as either kinetic or
potential
• energy can be converted from one form to
another
• when matter undergoes a chemical or
physical change, the amount of energy in
the matter changes as well
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Energy of Matter - Kinetic
• kinetic energy is energy of motion
motion of the atoms, molecules, and
subatomic particles
thermal (heat) energy is a form of kinetic
energy because it is caused by molecular
motion
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Energy of Matter - Potential
• potential energy is energy that is stored in
the matter
due to the composition of the matter and its
position in the universe
chemical potential energy arises from
electrostatic forces between atoms, molecules,
and subatomic particles
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Conversion of Energy
• you can interconvert kinetic energy and
potential energy
• whatever process you do that converts
energy from one type or form to another, the
total amount of energy remains the same
Law of Conservation of Energy
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Spontaneous Processes
• materials that possess high potential
•
energy are less stable
processes in nature tend to occur on
their own when the result is material(s)
with lower total potential energy
 processes that result in materials with
higher total potential energy can occur, but
generally will not happen without input of
energy from an outside source
• when a process results in materials
with less potential energy at the end
than there was at the beginning, the
difference in energy is released into the
environment
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Potential to Kinetic Energy
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Standard Units of Measure
The Standard Units
• Scientists have agreed on a set of
international standard units for comparing
all our measurements called the SI units
Système International = International System
Quantity
length
mass
time
temperature
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Unit
meter
kilogram
second
kelvin
Symbol
m
kg
s
K
42
Length
• Measure of the two-dimensional distance an object covers
 often need to measure lengths that are very long (distances
between stars) or very short (distances between atoms)
• SI unit = meter
 About 3.37 inches longer than a yard
1 meter = one ten-millionth the distance from the North Pole to
the Equator = distance between marks on standard metal rod =
distance traveled by light in a specific period of time
• Commonly use centimeters (cm)
 1 m = 100 cm
 1 cm = 0.01 m = 10 mm
 1 inch = 2.54 cm (exactly)
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Mass
• Measure of the amount of matter
present in an object
 weight measures the gravitational pull on
an object, which depends on its mass
• SI unit = kilogram (kg)
 about 2 lbs. 3 oz.
• Commonly measure mass in grams (g)
or milligrams (mg)
 1 kg = 2.2046 pounds, 1 lbs. = 453.59 g
 1 kg = 1000 g = 103 g
 1 g = 1000 mg = 103 mg
 1 g = 0.001 kg = 10-3 kg
 1 mg = 0.001 g = 10-3 g
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Time
• measure of the duration of an event
• SI units = second (s)
• 1 s is defined as the period of time it
takes for a specific number of
radiation events of a specific
transition from cesium-133
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Temperature
• measure of the average amount of
kinetic energy
 higher temperature = larger average
kinetic energy
• heat flows from the matter that has
high thermal energy into matter that
has low thermal energy
 until they reach the same temperature
 heat is exchanged through molecular
collisions between the two materials
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Temperature Scales
• Fahrenheit Scale, °F
 used in the U.S.
• Celsius Scale, °C
 used in all other countries
• Kelvin Scale, K
 absolute scale
 no negative numbers
 directly proportional to
average amount of kinetic
energy
 0 K = absolute zero
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Fahrenheit vs. Celsius
• a Celsius degree is 1.8 times larger than a
Fahrenheit degree
• the standard used for 0° on the Fahrenheit
scale is a lower temperature than the
standard used for 0° on the Celsius scale
C 
F - 32
1.8
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Kelvin vs. Celsius
• the size of a “degree” on the Kelvin scale is
the same as on the Celsius scale
though technically, we don’t call the divisions
on the Kelvin scale degrees; we called them
kelvins!
so 1 kelvin is 1.8 times larger than 1°F
• the 0 standard on the Kelvin scale is a much
lower temperature than on the Celsius scale
K  C  273.15
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Example 1.2 Convert 40.00 °C into K and °F
•
•
•
Find the equation that relates
Given:
the given quantity to the
Find:
quantity you want to find
Equation:
Since the equation is solved
for the quantity you want to
find, substitute and compute
40.00 °C
K
K = °C + 273.15
K = °C + 273.15
K = 40.00 + 273.15
K = 313.15 K
Find the equation that relates
Given:
the given quantity to the
Find:
quantity you want to find
Equation:
40.00 °C
°F
C 
 F - 32 
1.8
•
•
Solve the equation for the
quantity you want to find
Substitute and compute
1 . 8   C   F - 32 
1 . 8   C  32   F
1 . 8  40 . 00  32   F
104.00  F   F
Related Units in the
SI System
• All units in the SI system are related to the
standard unit by a power of 10
• The power of 10 is indicated by a prefix
multiplier
• The prefix multipliers are always the same,
regardless of the standard unit
• Report measurements with a unit that is close to
the size of the quantity being measured
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Common Prefix Multipliers in the
SI System
Prefix
Symbol
Decimal
Equivalent
Power of 10
1,000,000
Base x 106
1,000
Base x 103
mega-
M
kilo-
k
deci-
d
0.1
Base x 10-1
centi-
c
0.01
Base x 10-2
milli-
m
0.001
Base x 10-3
micro-
m or mc
0.000 001
Base x 10-6
nano-
n
0.000 000 001 Base x 10-9
pico
p
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0.000 000 000 001
Base x 10-12
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• Derived unit
Volume
 any length unit cubed
• Measure of the amount of space occupied
• SI unit = cubic meter (m3)
• Commonly measure solid volume in cubic
centimeters (cm3)
 1 m3 = 106 cm3
 1 cm3 = 10-6 m3 = 0.000001 m3
• Commonly measure liquid or gas volume
in milliliters (mL)
 1 L is slightly larger than 1 quart
 1 L = 1 dm3 = 1000 mL = 103 mL
 1 mL = 0.001 L = 10-3 L
 1 mL = 1 cm3
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Common Units and Their Equivalents
Length
1 kilometer (km)
1 meter (m)
1 meter (m)
1 foot (ft)
1 inch (in.)
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=
=
=
=
=
0.6214 mile (mi)
39.37 inches (in.)
1.094 yards (yd)
30.48 centimeters (cm)
2.54 centimeters (cm) exactly
54
Common Units and Their Equivalents
Mass
1 kilogram (km) = 2.205 pounds (lb)
1 pound (lb) = 453.59 grams (g)
1 ounce (oz) = 28.35 grams (g)
Volume
1 liter (L)
1 liter (L)
1 liter (L)
1 U.S. gallon (gal)
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=
=
=
=
1000 milliliters (mL)
1000 cubic centimeters (cm3)
1.057 quarts (qt)
3.785 liters (L)
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Density
Mass & Volume
• two main physical properties of matter
• mass and volume are extensive properties
the value depends on the quantity of matter
extensive properties cannot be used to identify
what type of matter something is
if you are given a large glass containing 100 g of a
clear, colorless liquid and a small glass containing
25 g of a clear, colorless liquid - are both liquids the
same stuff?
• even though mass and volume are
individual properties, for a given type of
matter they are related to each other!
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Mass vs. Volume of Brass
Mass
grams
20
Volume
3
cm
2.4
32
3.8
40
4.8
50
6.0
100
11.9
150
17.9
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Volume vs. Mass of Brass
y = 8.38x
160
140
120
Mass, g
100
80
60
40
20
0
0.0
2.0
4.0
6.0
8.0
10.0
12.0
14.0
16.0
18.0
Volume, cm3
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Density
• Ratio of mass:volume is an intensive property
value independent of the quantity of matter
• Solids = g/cm3
1 cm3 = 1 mL
Density 
Mass
Volume
• Liquids = g/mL
• Gases = g/L
• Volume of a solid can be determined by water
displacement – Archimedes Principle
• Density : solids > liquids >>> gases
except ice is less dense than liquid water!
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Density 
Mass
Density
Volume
• For equal volumes, denser
•
•
object has larger mass
For equal masses, denser
object has smaller volume
Heating an object generally
causes it to expand, therefore
the density changes with
temperature
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Example 1.3 Decide if a ring with a mass of 3.15 g
that displaces 0.233 cm3 of water is platinum
•
Find the equation that relates
the given quantity to the
quantity you want to find
Given:
Find:
Equation:
•
•
Since the equation is solved
for the quantity you want to
find, and the units are
correct, substitute and
compute
Compare to accepted value
of the intensive property
mass = 3.15 g
volume = 0.233 cm3
density, g/cm3
Density 
Mass
Volume
d 
m
V

3 . 15 g
0.233 cm
d  13.5 g/cm
3
3
Density of platinum =
21.4 g/cm3
therefore not
platinum
Measurement
and Significant Figures
What Is a Measurement?
• quantitative observation
• comparison to an agreedupon standard
• every measurement has a
number and a unit
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A Measurement
• the unit tells you what standard you are
comparing your object to
• the number tells you
1. what multiple of the standard the object measures
2. the uncertainty in the measurement
• scientific measurements are reported so that
every digit written is certain, except the last
one which is estimated
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Estimating the Last Digit
• for instruments marked with a scale, you get
the last digit by estimating between the marks
 if possible
• mentally divide the space into 10 equal
spaces, then estimate how many spaces over
the indicator mark is
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Significant Figures
• the non-place-holding digits in a
reported measurement are called
significant figures
 some zeros in a written number are
only there to help you locate the
decimal point
• significant figures tell us the range
of values to expect for repeated
measurements
 the more significant figures there are
in a measurement, the smaller the
range of values is
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12.3 cm
has 3 sig. figs.
and its range is
12.2 to 12.4 cm
12.30 cm
has 4 sig. figs.
and its range is
12.29 to 12.31 cm
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Counting Significant Figures
1) All non-zero digits are significant
 1.5 has 2 sig. figs.
2) Interior zeros are significant
 1.05 has 3 sig. figs.
3) Leading zeros are NOT significant
 0.001050 has 4 sig. figs.
 1.050 x 10-3
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Counting Significant Figures
4) Trailing zeros may or may not be significant
1) Trailing zeros after a decimal point are significant

1.050 has 4 sig. figs.
2) Zeros at the end of a number without a written
decimal point are ambiguous and should be avoided
by using scientific notation
 if 150 has 2 sig. figs. then 1.5 x 102
 but if 150 has 3 sig. figs. then 1.50 x 102
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Significant Figures and Exact Numbers
• Exact numbers have an unlimited number of
significant figures
• A number whose value is known with
complete certainty is exact
from counting individual objects
from definitions
1 cm is exactly equal to 0.01 m
from integer values in equations
 in the equation for the radius of a circle, the 2 is exact
radius of a circle = diameter of a circle
2
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Example 1.5 Determining the Number of
Significant Figures in a Number
How many significant figures are in each of the following?
0.04450 m
4 sig. figs.; the digits 4 and 5, and the trailing 0
5.0003 km
5 sig. figs.; the digits 5 and 3, and the interior 0’s
10 dm = 1 m
infinite number of sig. figs., exact numbers
1.000 × 105 s
4 sig. figs.; the digit 1, and the trailing 0’s
0.00002 mm
1 sig. figs.; the digit 2, not the leading 0’s
10,000 m
Ambiguous, generally assume 1 sig. fig.
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Multiplication and Division with
Significant Figures
• when multiplying or dividing measurements with
significant figures, the result has the same number of
significant figures as the measurement with the
fewest number of significant figures
5.02 ×
89,665 × 0.10 = 45.0118 = 45
3 sig. figs.
5 sig. figs.
5.892 ÷
4 sig. figs.
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2 sig. figs.
2 sig. figs.
6.10 = 0.96590 = 0.966
3 sig. figs.
3 sig. figs.
73
Addition and Subtraction with
Significant Figures
• when adding or subtracting measurements with
significant figures, the result has the same number of
decimal places as the measurement with the fewest
number of decimal places
5.74 +
0.823 +
2.651 = 9.214 = 9.21
2 dec. pl.
4.8
1 dec. pl
3 dec. pl.
-
3.965
3 dec. pl.
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3 dec. pl.
=
0.835 =
2 dec. pl.
0.8
1 dec. pl.
74
Rounding
•
1.
2.
•
when rounding to the correct number of significant
figures, if the number after the place of the last
significant figure is
0 to 4, round down
 drop all digits after the last sig. fig. and leave the last
sig. fig. alone
 add insignificant zeros to keep the value if necessary
5 to 9, round up
 drop all digits after the last sig. fig. and increase the
last sig. fig. by one
 add insignificant zeros to keep the value if necessary
to avoid accumulating extra error from rounding, round
only at the end, keeping track of the last sig. fig. for
intermediate calculations
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Rounding
• rounding to 2 significant figures
• 2.34 rounds to 2.3
because the 3 is where the last sig. fig. will be
and the number after it is 4 or less
• 2.37 rounds to 2.4
because the 3 is where the last sig. fig. will be
and the number after it is 5 or greater
• 2.349865 rounds to 2.3
because the 3 is where the last sig. fig. will be
and the number after it is 4 or less
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Rounding
• rounding to 2 significant figures
• 0.0234 rounds to 0.023 or 2.3 × 10-2
because the 3 is where the last sig. fig. will be
and the number after it is 4 or less
• 0.0237 rounds to 0.024 or 2.4 × 10-2
because the 3 is where the last sig. fig. will be
and the number after it is 5 or greater
• 0.02349865 rounds to 0.023 or 2.3 × 10-2
because the 3 is where the last sig. fig. will be
and the number after it is 4 or less
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Rounding
• rounding to 2 significant figures
• 234 rounds to 230 or 2.3 × 102
because the 3 is where the last sig. fig. will be
and the number after it is 4 or less
• 237 rounds to 240 or 2.4 × 102
because the 3 is where the last sig. fig. will be
and the number after it is 5 or greater
• 234.9865 rounds to 230 or 2.3 × 102
because the 3 is where the last sig. fig. will be
and the number after it is 4 or less
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Both Multiplication/Division and
Addition/Subtraction with Significant
Figures
• when doing different kinds of operations with
measurements with significant figures, do whatever
is in parentheses first, evaluate the significant
figures in the intermediate answer, then do the
remaining steps
3.489 × (5.67 – 2.3) =
2 dp
1 dp
3.489
×
3.37
=
12
4 sf
1 dp & 2 sf
2 sf
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79
Example 1.6 Perform the following calculations to
the correct number of significant figures
a)
b)
1 . 10  0 . 5120  4 . 0015  3 . 4555
0 . 355
 105 . 1
 100 . 5820
c)
4 . 562  3 . 99870   452 . 6755  452 . 33 
d)
14 . 84  0 . 55   8 . 02
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80
Example 1.6 Perform the following calculations to
the correct number of significant figures
a)
b)
1 . 10  0 . 5120  4 . 0015  3 . 4555  0 . 65219  0 . 652
0 . 355
 105 . 1
 100 . 5820
4 . 8730  4.9
c)
4 . 562  3 . 99870   452 . 6755  452 . 33   52 . 79904  53
d)
14 . 84  0 . 55   8 . 02
Tro, Chemistry: A Molecular Approach
 0 . 142  0 . 1
81
Precision
and Accuracy
Uncertainty in Measured Numbers
• uncertainty comes from limitations of the instruments
•
•
•
used for comparison, the experimental design, the
experimenter, and nature’s random behavior
to understand how reliable a measurement is we need
to understand the limitations of the measurement
accuracy is an indication of how close a measurement
comes to the actual value of the quantity
precision is an indication of how reproducible a
measurement is
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83
Precision
• imprecision in measurements is caused by
random errors
errors that result from random fluctuations
no specific cause, therefore cannot be corrected
• we determine the precision of a set of
measurements by evaluating how far they are
from the actual value and each other
• even though every measurement has some
random error, with enough measurements
these errors should average out
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84
Accuracy
• inaccuracy in measurement caused by
systematic errors
errors caused by limitations in the instruments or
techniques or experimental design
can be reduced by using more accurate instruments,
or better technique or experimental design
• we determine the accuracy of a measurement by
evaluating how far it is from the actual value
• systematic errors do not average out with
repeated measurements because they
consistently cause the measurement to be either
too high or too low
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85
Accuracy vs. Precision
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86
Solving
Chemical
Problems
Equations &
Dimensional Analysis
Units
• Always write every number with its
associated unit
• Always include units in your calculations
you can do the same kind of operations on units
as you can with numbers
cm × cm = cm2
cm + cm = cm
cm ÷ cm = 1
using units as a guide to problem solving is
called dimensional analysis
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88
Problem Solving and
Dimensional Analysis
• Many problems in chemistry involve using
•
relationships to convert one unit of measurement to
another
Conversion factors are relationships between two units
 May be exact or measured
• Conversion factors generated from equivalence
statements
 e.g., 1 inch = 2.54 cm can give
2.54cm
1in
Tro, Chemistry: A Molecular Approach
or
1in
2.54cm
89
Problem Solving and
Dimensional Analysis
• Arrange conversion factors so given unit cancels
Arrange conversion factor so given unit is on the
bottom of the conversion factor
• May string conversion factors
So we do not need to know every relationship, as
long as we can find something else the given and
desired units are related to
given unit 
desired unit
 desired unit
given unit
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90
Conceptual Plan
• a conceptual plan is a visual outline that shows
the strategic route required to solve a problem
• for unit conversion, the conceptual plan focuses
on units and how to convert one to another
• for problems that require equations, the
conceptual plan focuses on solving the equation
to find an unknown value
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91
Concept Plans and
Conversion Factors
• Convert inches into centimeters
1) Find relationship equivalence: 1 in = 2.54 cm
2) Write concept plan
in
cm
3) Change equivalence into conversion factors
with starting units on the bottom
2.54 cm
1 in
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92
Systematic Approach
• Sort the information from the problem
 identify the given quantity and unit, the quantity and unit
you want to find, any relationships implied in the problem
• Design a strategy to solve the problem
 Concept plan
 sometimes may want to work backwards
 each step involves a conversion factor or equation
• Apply the steps in the concept plan
 check that units cancel properly
 multiply terms across the top and divide by each bottom term
• Check the answer
 double check the set-up to ensure the unit at the end is the one
you wished to find
 check to see that the size of the number is reasonable
 since centimeters are smaller than inches, converting inches to
centimeters should result in a larger number
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93
Example 1.7 Convert 1.76 yd. to centimeters
•
•
Sort
information
Strategize
Given:
Find:
Concept Plan:
1.76 yd
length, cm
yd
Relationships:
•
•
•
Follow the
concept plan to
solve the
problem
Sig. figs. and
round
Check
m
cm
1 yd = 1.094 m
1 m = 100 cm
Solution:
1.79 yd 
1m
1.094 yd
Round:

100 cm
 160.8775 cm
1m
160.8775 cm = 161 cm
Check: Units & magnitude are correct
Practice – Convert 30.0 mL to quarts
(1 L = 1.057 qt)
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95
Convert 30.0 mL to quarts
•
•
Sort
information
Strategize
Given:
Find:
Concept Plan:
30.0 mL
volume, qts
mL
Relationships:
•
•
•
Follow the
concept plan
to solve the
problem
Sig. figs. and
round
Check
L
qt
1 L = 1.057 qt
1 L = 1000 mL
Solution:
1L
30.0 mL 

1000 mL
Round:
Check:
1.057 qt
 0.03171 qt
1L
0.03171 qt = 0.0317 qt
Units & magnitude are correct
Concept Plans for
Units Raised to Powers
• Convert cubic inches into cubic centimeters
1) Find relationship equivalence: 1 in = 2.54 cm
2) Write concept plan
in3
cm3
3) Change equivalence into conversion factors
with given unit on the bottom
3
3
3
2.54 cm
16.4 cm
 2.54 cm 


 
3
3
3
1
in
1 in
1 in


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97
Example 1.9 Convert 5.70 L to cubic inches
•
•
Sort
information
Strategize
Given:
Find:
Concept Plan:
Relationships:
•
•
•
Follow the
concept plan
to solve the
problem
Sig. figs. and
round
Check
5.70 L
volume, in3
L
cm3
mL
in3
1 mL = 1 cm3, 1 mL = 10-3 L
1 cm = 2.54 in
Solution:
5.70 L 
1 mL
10
-3
 347.835 in
L

1 cm
3
1 mL

1 in 3
2.54 cm 3
3
Round:
347.835 in3 = 348 in3
Check:
Units & magnitude are correct
Practice 1.9 How many cubic
centimeters are there in 2.11 yd3?
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99
Practice 1.9 Convert 2.11 yd3 to cubic centimeters
•
•
Sort
information
Strategize
Given:
Find:
Concept Plan:
Relationships:
•
•
•
Follow the
concept plan
to solve the
problem
Sig. figs. and
round
Check
Solution:
2.11 yd3
volume, cm3
yd3
in3
1 yd = 36 in
1 in = 2.54 cm
3
3




36
in
2
.54
cm
3
2.11 yd 

3
1 yd 
1 in 3
 1613210.75 cm
Round:
Check:
cm3
3
1613210.75 cm3
= 1.61 x 106 cm3
Units & magnitude are correct
Density as a Conversion Factor
•
can use density as a conversion factor
between mass and volume!!
 density of H2O = 1.0 g/mL \ 1.0 g H2O = 1 mL H2O
 density of Pb = 11.3 g/cm3 \ 11.3 g Pb = 1 cm3 Pb
How much does 4.0 cm3 of lead weigh?
4.0 cm3 Pb x
Tro, Chemistry: A Molecular Approach
11.3 g Pb
1 cm3 Pb
= 45 g Pb
101
Example 1.10 What is the mass in kg of 173,231 L
of jet fuel whose density is 0.738 g/mL?
•
•
Sort
information
Strategize
Given:
Find:
Concept Plan:
Relationships:
•
•
•
Follow the
concept plan
to solve the
problem
Sig. figs. and
round
Check
173,231 L
density = 0.738 g/mL
mass, kg
L
mL
g
kg
1 mL = 0.738 g, 1 mL = 10-3 L
1 kg = 1000 g
Solution:
173,231 L 
1 mL
10
-3
L

0.738 g
1 mL

1 kg
1000 g
5
 1.33 10 kg
Round:
Check:
1.33 x 105 kg
Units & magnitude are correct
Order of Magnitude Estimations
• using scientific notation
• focus on the exponent on 10
• if the decimal part of the number is less than 5,
just drop it
• if the decimal part of the number is greater than
5, increase the exponent on 10 by 1
• multiply by adding exponents, divide by
subtracting exponents
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103
Estimate the Answer
• Suppose you count 1.2 x 105 atoms per second
for a year. How many would you count?
1 s = 1.2 x 105  105 atoms
1 minute = 6 x 101  102 s
1 hour = 6 x 101  102 min
1 day = 24  101 hr
1 yr = 365  102 days
2
1 yr 
10 days
1

1 yr
12
 10
10 hr
1 day
2

10 min
1 hr
2

10 s
1 min
5

10 atoms
1s
atoms
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Problem Solving with Equations
• When solving a problem involves using an
equation, the concept plan involves being given
all the variables except the one you want to find
• Solve the equation for the variable you wish to
find, then substitute and compute
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105
Using Density in Calculations
Concept Plans:
Density 
Mass
m, V
D
m, D
V
V, D
m
Volume
Volume 
Mass
Density
Mass  Density  Volume
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106
Example 1.12 Find the density of a metal cylinder
with mass 8.3 g, length 1.94 cm, and radius 0.55 cm
•
•
Sort
information
Strategize
Given:
Find:
Concept Plan:
m = 8.3 g
l = 1.94 cm, r = 0.55 cm
density, g/cm3
l, r
V
•
•
Follow the
concept plan
to solve the
problem
Sig. figs. and
round
Check
Solution:
V = π (0.55 cm)2 (1.94 cm)
V = 1.8436 cm3
d 
8.3 g
1. 8 436 cm
d  4 . 5 g/cm
Check:
d
V = π r2 l
d = m/V
Relationships:
•
m, V
3
 4 . 50206 g/cm
3
Units & magnitude OK
3